Question

The molar mass of carbon dioxide is 0.045kg/mol. Calculate the density of the gas when the temperature is 273K and the pressure is 120000Pa..?

Answer 1

`rho_"carbon dioxide"=2.4*g*L^-1`

Explanation:

We assume ideality, and we know that `PV=nRT`

And so `P/(RT)=n/V=("mass"/"molar mass")/V`

And so `P/(RT)xx"molar mass"=underbrace("mass"/V)_("density", rho)`...

And so we plug in the numbers....

`rho_"carbon dioxide"=(120*kPa)/((273*Kxx8.31*L* kPa)/(K*mol))xx0.045*kg*mol^-1=2.38xx10^-3*kg*L^-1`

Answer 2

The density is `"2.3 kg/m"^3`.

Explanation:

The molar mass of `"CO"_2` is 0.044 kg/mol.

We can use the Ideal Gas Law to determine its density.

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

Since `n = m/M`, we can substitute this to get

`pV = (m/M)RT`

We can rearrange this to

`pM = m/VRT`

But `"density"= "mass"/"volume"` or `color(brown)(bar(ul(|color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "`

∴ `pM = ρRT`

and

`color(brown)(bar(ul(|color(white)(a/a)ρ = (pM)/(RT)color(white)(a/a)|)))" "`

In your problem,

`p = "120 000 Pa"`
`M = "0.040 g/mol"`
`R = "8.314 Pa·m"^3·"K"^"-1""mol"^"-1"`
`T ="273K"`

∴ `ρ = ("120 000" color(red)(cancel(color(black)("Pa"))) × 0.044 color(white)(l) "kg"·color(red)(cancel(color(black)("mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("Pa")))·"m"^3color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273color(red)(cancel(color(black)("K")))) = "2.3 kg/m"^3`